Gauss-Lucas Theorem

Theorem

Let $P$ be a (nonconstant) polynomial with complex coefficients.

Then all zeros of its derivative $P'$ belong to the convex hull of the set of zeros of $P$.

Proof

Over the complex numbers, $P$ is a product of prime factors:

$\displaystyle P(z)= \alpha \prod_{i \mathop = 1}^n (z-a_i)$

where:

the complex numbers $a_1, a_2, \ldots, a_n$ are the – not necessary distinct – zeros of the polynomial $P$

the complex number \(\alpha\) is the leading coefficient of $P$

$n$ is the degree of $P$.

Let $z$ be any complex number for which $P(z) \neq 0$.

Then we have for the logarithmic derivative:

$\displaystyle \frac {P' (z)} {P(z)} = \sum_{i \mathop = 1}^n \frac 1 {z-a_i}$

In particular, if $z$ is a zero of $P'$ and $P(z) \neq 0$, then:

$\displaystyle \sum_{i \mathop = 1}^n \frac 1 {z-a_i} = 0$

or:

$\displaystyle \sum_{i \mathop = 1}^n \frac {\overline z - \overline {a_i} } {\vert z - a_i \vert^2} = 0$

This may also be written as

$\displaystyle \left(\sum_{i \mathop = 1}^n \frac 1 {\vert z - a_i \vert^2}\right) \overline z = \left(\sum_{i \mathop = 1}^n \frac 1 {\vert z - a_i\vert^2} \overline{a_i}\right)$

Taking their conjugates, we see that $z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $a_i$ (with different mass assigned on each root whose weights collectively sum to 1).

If $P (z) = P' (z)=0$, then $z = 1 \cdot z + 0 \cdot a_i$, and is still a convex combination of the roots of $P$.

$\blacksquare$

Source of Name

This entry was named for Carl Friedrich Gauss and Édouard Lucas.